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24 Jul 2011, 02:40

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

87%(02:05) correct
13%(02:30) wrong based on 23 sessions

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The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=

a)4b)5c)6d)7e)8

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10. M=28, so x = 7

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24 Jul 2011, 04:58

tracyyahoo wrote:

The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=

a)4b)5c)6d)7e)8

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10. M=28, so x = 7

why x = 7 can be concluded from M=28?

sum of first x terms where 1<x<10 is x(x+1)/2 --- Eq 1Now we know that sum of terms in Arithmetic Progression is n/2[2a +d(n-1)]where n = number of terms; here n= 10-x.a = first number of the seq. here first term will be x+1d = difference between 2 consecutive numbers; 1 in this case

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Both the replies given so far are correct, but I think there's a better way to approach this - use the answers. The answers are there to help you - you should always read them as part of the question.

One way to solve this problem would be to guess and check, then adjust your guess based on whether it was too high or too low. In these situation you should pick C as your first guess and check number. Because the GMAT always puts numerical answers in order, picking C will either get you the correct answer or eliminate 3 choices immediately. Then you only have to check one more answer

Guess C: 1 + 2 + 3 + 4 + 5 + 6 = 21 and 7 + 8 + 9 + 10 = 34. Too low

Now, obviously A and B can't be the answers, so we'll check D, and if that doesn't work E will be our answer.

Guess D: 21 + 7 = 28 and 34 - 7 = 27. Correct

Note that I used calculations I'd already done for my first guess in my second guess.

Most questions on the GMAT can be answered with a little common sense - students spending hours relearning algebra and geometry that they've forgotten since high school or college often overlook this approach.

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25 Jul 2011, 09:29

I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently,Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)

We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10

The sum of the first x terms = x(1 + x)/2

Now we have "10 - x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10 - x )(x + 1 + 10)/2 = (10 - x)(x + 11)/2

But x(x + 1)/2 - 1= (10 - x)(x + 11)/2

Now solving above equation we get a quadratic equation x^2 + x - 56 = 0 and we can factor it as

(x + 8) (x - 7) = 0Hence x = -8, 7 Reject -8 as it is not in page numbers and our answer is x = 7 which is given by choice d.
_________________

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25 Jul 2011, 17:26

Thank u . ur post is most comprehensable.

uote="2ndgrademath"]I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently,Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)

We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10

The sum of the first x terms = x(1 + x)/2

Now we have "10 - x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10 - x )(x + 1 + 10)/2 = (10 - x)(x + 11)/2

But x(x + 1)/2 - 1= (10 - x)(x + 11)/2

Now solving above equation we get a quadratic equation x^2 + x - 56 = 0 and we can factor it as

(x + 8) (x - 7) = 0Hence x = -8, 7 Reject -8 as it is not in page numbers and our answer is x = 7 which is given by choice d.[/quote]

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25 Jul 2011, 21:09

why x(x-1)/2 - 1? I don"t follow this. Pls help me to explaim this

2ndgrademath wrote:

I want to use the AP sums, same as Sudhanshuacharya, but explain a bit differently,Use of the formal to sum the AP with first term "a", last term "l" and number of terms "n" is given as shown next, = n/2 (a + l)

We can sum the series as " 1 + 2 + ......+ x + (x+1) + .....+10

The sum of the first x terms = x(1 + x)/2

Now we have "10 - x" more terms starting with "x+1" and last term is 10, we can find the sum of these too; as = (10 - x )(x + 1 + 10)/2 = (10 - x)(x + 11)/2

But x(x + 1)/2 - 1= (10 - x)(x + 11)/2

Now solving above equation we get a quadratic equation x^2 + x - 56 = 0 and we can factor it as

(x + 8) (x - 7) = 0Hence x = -8, 7 Reject -8 as it is not in page numbers and our answer is x = 7 which is given by choice d.

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The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=

a)4b)5c)6d)7e)8

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10. M=28, so x = 7

why x = 7 can be concluded from M=28?

The moment I look at the question, I find myself jumping to option C to see if it works. Notice that adding numbers between 1 to 10 is really easy and quick. I wouldn't bother using Algebra in this question. But if we change this to pages numbered from 1 to 100 or something, I wouldn't bother with the options and go to Algebra instead. So knowing both the techniques is important. Let me discuss both of them.

Case 1: Using OptionsIf I add 1 to 6, I get 6*7/2 = 21Adding 7+8+9+10 gives me 34 (You can simply add these OR say that the sum will be (3+2+1) less than 40 i.e. 6 less than 40 OR multiply the average (8.5) of the numbers by 4 i.e. 8.5*4 = 8*4 + .5*4 = 34)Nope. We need to go to D and do the same there. We see that it works.

Note: Here the solution came easily as 56=7*8. But you might have something like n*(n+1) < 45 normally in a question ... then you need to just do some mental math to find largest n, such that n*(n+1) < 45... This is how I would start thinking... 5*6 = 30, 6*7 = 42, so n = 6. So be prepared to do some easy mental math in the Test. From all the tests I have given, I have seen at-least one problem that has Sum of an arithmetic progression (S(n) is the simplest form of Arithmetic pregression) combined with inequality.

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28 Jul 2011, 04:52

VeritasPrepKarishma wrote:

tracyyahoo wrote:

The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=

a)4b)5c)6d)7e)8

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10. M=28, so x = 7

why x = 7 can be concluded from M=28?

The moment I look at the question, I find myself jumping to option C to see if it works. Notice that adding numbers between 1 to 10 is really easy and quick. I wouldn't bother using Algebra in this question. But if we change this to pages numbered from 1 to 100 or something, I wouldn't bother with the options and go to Algebra instead. So knowing both the techniques is important. Let me discuss both of them.

Case 1: Using OptionsIf I add 1 to 6, I get 6*7/2 = 21Adding 7+8+9+10 gives me 34 (You can simply add these OR say that the sum will be (3+2+1) less than 40 i.e. 6 less than 40 OR multiply the average (8.5) of the numbers by 4 i.e. 8.5*4 = 8*4 + .5*4 = 34)Nope. We need to go to D and do the same there. We see that it works.

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If we add 1 to 7, then we get 7*8/2=28Adding 8+9+10, we get 27. ( average = 27/3=9)Then how we get 7. Is the fact that 10*11/2=55 and 28+27=55.

The condition that the answer should satisfy is this: If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x

If x = 7, the sum of page number up to and including x is 1+2+3..+6+7 = 7*8/2 = 28 (The sum of n consecutive numbers starting from 1 is n(n+1)/2)The sum of the page numbers following page number x = 8+9+10 = 27Since 28 is one more than 27, our condition is satisfied. So answer is 7.
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09 Aug 2011, 20:33

tracyyahoo wrote:

The pages of a report are numbered consecutively from 1 to 10. If the sum of the page numbers up to and including page number x of the report is equal to one more than the sum of the page numbers following page number x, then x=

a)4b)5c)6d)7e)8

I have no idea how to solve this ,pls help~~~

the explanation says that suppose the sum of x is m, then 2m-1=1+2+3+.......+10. M=28, so x = 7

why x = 7 can be concluded from M=28?

Another approach:

Sum of the page numbers from 1 to 10 = 55

The sum of the page numbers up to and including page number x of the report = aThe sum of the page numbers following page number x = b